Trong các hàm số sau, hàm số nào tăng trên khoảng $\left( { - 1;0} \right)$?

Lấy $- 1 < {x_1} < {x_2} < 0$ thì ${x_2} - {x_1} > 0$ ta có:

$T = \dfrac{{f\left( {{x_2}} \right) - f\left( {{x_1}} \right)}}{{{x_2} - {x_1}}} = \dfrac{{{x_2} - {x_1}}}{{{x_2} - {x_1}}} = 1 > 0,\forall {x_1},{x_2} \in \left( { - 1;0} \right)$  nên đáp án A đúng.

$T = \dfrac{{f\left( {{x_2}} \right) - f\left( {{x_1}} \right)}}{{{x_2} - {x_1}}} = \dfrac{{\dfrac{1}{{{x_2}}} - \dfrac{1}{{{x_1}}}}}{{{x_2} - {x_1}}} = \dfrac{{{x_1} - {x_2}}}{{{x_1}{x_2}\left( {{x_2} - {x_1}} \right)}} =  - \dfrac{1}{{{x_1}{x_2}}} < 0,\forall {x_1},{x_2} \in \left( { - 1;0} \right)$ nên B sai.

$T = \dfrac{{f\left( {{x_2}} \right) - f\left( {{x_1}} \right)}}{{{x_2} - {x_1}}} = \dfrac{{\left| {{x_2}} \right| - \left| {{x_1}} \right|}}{{{x_2} - {x_1}}} = \dfrac{{ - {x_2} + {x_1}}}{{{x_2} - {x_1}}} =  - 1 < 0,\forall {x_1},{x_2} \in \left( { - 1;0} \right)$ nên C sai.

$T = \dfrac{{f\left( {{x_2}} \right) - f\left( {{x_1}} \right)}}{{{x_2} - {x_1}}} = \dfrac{{x_2^2 - x_1^2}}{{{x_2} - {x_1}}} = {x_2} + {x_1} < 0,\forall {x_1},{x_2} \in \left( { - 1;0} \right)$ nên D sai.